Two Dimensional Analytical Geometry Practical Problems and Solutions :
Here we are going to see some practical problems in two dimensional analytical geometry.
Question 1 :
A family is using Liquefied petroleum gas (LPG) of weight 14.2 kg for consumption. (Full weight 29.5 kg includes the empty cylinders tare weight of 15.3 kg.). If it is use with constant rate then it lasts for 24 days. Then the new cylinder is replaced (i) Find the equation relating the quantity of gas in the cylinder to the days. (ii) Draw the graph for first 96 days.
Solution :
From the given information, we may take two points
(0, 14.2) (24, 0)
(y−y_{1})/(y_{2}−y_{1}) = (x−x_{1})/(x_{2}-x_{1})
(y − 14.2)/(0 - 14.2) = (x−0)/(24 - 0)
(y − 14.2)/(- 14.2) = x/24
24(y - 14.2) = -14.2x
y - 14.2 = -14.2x/24
y = -(7.1/12)x + 14.2
Since we are replacing the cylinder for every 24 days once, we have to consider this as periodic function.
f(x) = f(x + 24)
Question 2 :
In a shopping mall there is a hall of cuboid shape with dimension 800 × 800 × 720 units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find (i)the minimum total length of the escalator. (ii) the heights at which the escalator changes its direction. (iii) the slopes of the escalator at the turning points.
Solution :
(i) OAB is a right triangle
OA^{2} = OB^{2} + AB^{2}
OA^{2} = 3200^{2} + 720^{2}
OA^{2} = 10240000 + 518400
= 10758400
= 3280 m
(ii) Using the points (0, 0) and (3200, 720) we may find the equation of the line
(y - 0)/(720 - 0) = (x - 0)/(3200 - 0)
y/720 = x/3200
y = 720x/3200
y = (9/40)x
The heights at which the escalator changes its direction
When x = 800 y = (9/40)(800) = 180 |
When x = 1600 y=(9/40)(1600) = 360 |
When x = 2400 y=(9/40)(2400) = 540 |
(iii) the slopes of the escalator at the turning points.
Slope = 9/40
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